Solutions of Homework problems. Resistive circuits. Problem 1 Use KVL and Ohms law to compute voltages v a and v b . . v 1. From Ohms law: v 1 =8k W* i 1 =8[V]. v 2 =2k W* i 2 =2[V] Form KVL: v a =5[V]v 2 =7[V] v b =15[V]v 1 v a =0[V]. +. . v 2. +. +. +. . .
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Problem 1
Use KVL and Ohms law to compute voltages va and vb .

v1
From Ohms law:
v1=8kW*i1=8[V]
v2=2kW*i2=2[V]
Form KVL:
va=5[V]v2=7[V]
vb=15[V]v1va=0[V]
+

v2
+
+
+


Problem 2
Write equations to compute voltages v1 and v2 , next find the current value of i1
From KCL:
50 mA=v1/40+(v1v2)/40
and
100 mA=v2/80+(v2v1)/40
i1
i1
v1
v2
Multiply first equation by 40:
2=v1+v1v2=2v1v2
From second equation:
8=v2+2(v2v1)=3v22v1 add both sides:
10=2v2 => v2=5 [V], v1=1+v2 /2=3.5[V]
i1= (v1v2)/40=1.5/40=37.5 [mA]
100 mA
50 mA
Problem 3: Find Thevenin and Norton equivalent circuit for the network shown.
I1
N1
N2
I2
vt
From KVL
_
Thevenin & Norton
RTh=1.33Ω
A
A
RTh=vt/Isc=1.33Ω
RTh=1.33Ω
In=4.5 A
Vt=6 V
Note: Negative vt indicates that the polarity is reversed and as a result this circuit has a negative resistance.
B
B
Norton Equivalent
Thevenin Equivalent
Problem 4: Find the current i and the voltage v across LED diode in the circuit shown on Fig. a) assuming that the diode characteristic is shown on Fig. b).
Draw load line. Intersection of load line and diode characteristic is the i and v across LED diode: v ≈ 1.02 V and i ≈ 7.5 mA.
Problem 5: Sketch i versus v to scale for each of the circuits shown below. Assume that the diodes are ideal and allow v to range from 10 V to +10 V.
i
+
v
_
(a)
2kΩ
Diode is on for v > 0 and R=2kΩ.
In a series connection voltages are added for each constant current
Problem 5: Sketch i versus v to scale for each of the circuits shown below. Assume that the diodes are ideal and allow v to range from 10 V to +10 V.
i
+
v
_
(a)
2kΩ
Resulting characteristics
Problem 5: Sketch i versus v to scale for each of the circuits shown below. Assume that the diodes are ideal and allow v to range from 10 V to +10 V.
i
1kΩ
+
v
_
(b)
+
_
Due to the presence of the 5V supply the diode conducts only for v > 5, R = 1kΩ
5V
First combine diode and resistance then add the voltage source
Problem 5: Sketch i versus v to scale for each of the circuits shown below. Assume that the diodes are ideal and allow v to range from 10 V to +10 V.
+
v
_
2kΩ
1kΩ
A
B
(c)
Diode B is on for v > 0 and R=1kΩ.
Diode A is on for v < 0 and R=2kΩ.
Problem 5: Sketch i versus v to scale for each of the circuits shown below. Assume that the diodes are ideal and allow v to range from 10 V to +10 V.
(d)
i
Diode D is on for v > 0 and R=1kΩ.
Diode C is on for v < 0 and R=0Ω.
+
v
_
D
C
1kΩ
_
+
_
Problem 6: Assuming ideal diodes sketch to scale the transfer characteristics (vo versus vin) for the circuit shown below.
1kΩ
+
vo
_
1kΩ
vin
3V
+
vo
_
1kΩ
vin
Case I: vin > 0
Both diodes are on, and act as short circuits. The equivalent circuit is shown here.
vo = vin
_
+
_
Problem 6: Assuming ideal diodes sketch to scale the transfer characteristics (vo versus vin) for the circuit shown below.
1kΩ
1kΩ
+
vo
_
+
vo
_
1kΩ
1kΩ
vin
vin
Case II: vin < 0
Both diodes are reverse biased and vo is the sum of the voltage drops across Zener diode and 1kΩ resistor.
3V
3V
_
+
_
Problem 6: Assuming ideal diodes sketch to scale the transfer characteristics (vo versus vin) for the circuit shown below.
Case II: vin < 0
Both diodes are reverse biased and vo is the sum of the voltage drops across Zener diode and 1kΩ resistor.
1kΩ
+
vo
_
1kΩ
vin
3V
1kΩ
Case IIa: 3V < vin < 0
vo = vin, because the current through Zener diode is zero, all negative voltage drop is across the Zener diode.
+
vo
_
1kΩ
vin
_
+
_
Problem 6: Assuming ideal diodes sketch to scale the transfer characteristics (vo versus vin) for the circuit shown below.
vo
1kΩ
+
vo
_
1
1kΩ
1
vin
vin
3V
3V
3V
Case IIb: vin < 3V
Excess voltage below 3V is dropped across the two resistors (1kW and 1kW), with
vo = (1/2)*(vin+3)3= vin/21.5 [V].
1
2